Solubility of Sn(OH) 2 was found to be 1.11 x 10 -9 M; calculate K sp of this compound.

Here we're told that the solubility of 10 2 hydroxide was found to be 1.11 times 10 to the negative 9 molar. From this, we need to calculate the solubility crop product constant or ksp of this compound. So, first of all, they're telling us the solubility of the entire ionic solid. So this represents our x. What we're gonna do now is, because we're talking about KSP, we're gonna write this compound as a solid and show how it breaks up into its ions. Here it breaks up into tin 2 plus 2 hydroxide ions. So here we have initial change equilibrium. Remember, in an ICE chart we ignore solids and liquids, and with KSP, your reactant is a solid. Now we're gonna say we have 0 of our products initially. We lose reactants to make products, so this is plus x. There's a 2 here, so plus 2 x, plus x, plus 2 x. Ksp equals products, so it equals Tin 2 times hydroxide ion concentration. Remember, this is gonna be squared because of the coefficient. So here we're going to say it equals x times 2x squared. So x times 4x squared, which equals 4x cubed. Now here, we have our x from the very beginning, so plug it in. Remember your order of operations. You're going to cube that that solubility first, then multiply that answer by 4. So let me just do that. Let me just cube that concentration inside. So cubing it gives me 1.367631 times 10 to the negative 27. Now multiply that by 4, so our KSP here would be 5.47 times 10 to the negative 27. So that would be the solubility product constant of Tin 2 hydroxide.

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18. Aqueous Equilibrium

Solubility Product Constant: Ksp

General Chemistry

18. Aqueous Equilibrium

Solubility Product Constant: Ksp

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Multiple Choice

Solubility of Sn(OH)₂ was found to be 1.11 x 10^-9 M; calculate K_sp of this compound.

5.47 × 10⁻²⁷

4.44 × 10⁻⁹

1.23 × 10⁻¹⁸

1.37 × 10⁻²⁷

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Verified step by step guidance

Understand the dissolution process of Sn(OH)2 in water. The compound dissociates into Sn²⁺ and OH⁻ ions. The balanced chemical equation for this process is: Sn(OH)₂(s) ⇌ Sn²⁺(aq) + 2OH⁻(aq).

Identify the expression for the solubility product constant (Ksp) for Sn(OH)₂. The Ksp expression is derived from the equilibrium concentrations of the ions: Ksp = [Sn²⁺][OH⁻]².

Recognize that the solubility of Sn(OH)₂ is given as 1.11 x 10⁻⁹ M, which represents the concentration of Sn²⁺ ions at equilibrium. Therefore, [Sn²⁺] = 1.11 x 10⁻⁹ M.

Determine the concentration of OH⁻ ions. Since each formula unit of Sn(OH)₂ produces two OH⁻ ions, the concentration of OH⁻ ions is twice the solubility of Sn(OH)₂: [OH⁻] = 2 x 1.11 x 10⁻⁹ M.

Substitute the concentrations into the Ksp expression: Ksp = (1.11 x 10⁻⁹) * (2 x 1.11 x 10⁻⁹)². Calculate the Ksp value using these concentrations.